1990 AMC 12 Problem 26

Question

Ten people form a circle. Each picks a number and tells it to the two neighbors adjacent to him in the circle. Then each person computes and announces the average of the numbers of his two neighbors. The figure shows the average announced by each person (not the original number the person picked). The number picked by the person who announced the average

6

was

1

5

6

10

not uniquely determined from the given information

Answer: A

Concepts:recurrence linear system cyclic arrangement

Difficulty rating: 2500

Small hint:

If

x_i

is a picked number and

a_i

the displayed average, then

x_{i-1}+x_{i+1}=2a_i

Big hint:

Start with two unknown adjacent picked numbers and propagate around the circle

Solution:

Index the displayed averages

a_0,a_1,\ldots,a_9

clockwise as

1,2,\ldots,10,

and let

x_i

be the corresponding picked numbers. Write

x_0=t,\ x_1=u.

From

x_{i-1}+x_{i+1}=2a_i,

successive values are

\begin{aligned} x_2&=4-t, &x_3&=6-u,\\ x_4&=4+t, &x_5&=4+u,\\ x_6&=8-t, &x_7&=10-u,\\ x_8&=8+t, &x_9&=8+u. \end{aligned}

The two closing equations give

t=6

and

u=-3.

Hence

x_5=4+u=1.

Thus the correct answer is A.